What is the solution for y^'+y=sin(x) ?

The solution for y^'+y=sin(x) is y=-(cos(x))/2+(sin(x))/2+(c_{1})/(e^x)

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y^'+y=sin(x)

y^{^{'}} + y = sin (x)

y = - \frac{cos ( x )}{2} + \frac{sin ( x )}{2} + \frac{c _{1}}{e ^{x}}

Solution steps

y^{'} (x) + y = sin (x)

Solve linear ODE:

y = - \frac{cos ( x )}{2} + \frac{sin ( x )}{2} + \frac{c _{1}}{e ^{x}}

y = - \frac{cos ( x )}{2} + \frac{sin ( x )}{2} + \frac{c _{1}}{e ^{x}}

n = 1 \sum \infty n

d er i v a t i v e f (x) = ln (4 x)

y^{^{''}} + y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 0

h \to 0 lim (\frac{2 ^{h} - 1}{h})

\frac{d}{d x} (x^{6})

What is the solution for y^'+y=sin(x) ?
The solution for y^'+y=sin(x) is y=-(cos(x))/2+(sin(x))/2+(c_{1})/(e^x)